JP2860405B2 - LED drive circuit - Google Patents

Description

Description: BACKGROUND OF THE INVENTION 1. Field of the Invention The present invention relates to an LED drive circuit including a booster circuit necessary for driving an LED (light emitting diode).

[Conventional technology]

For example, in a device that uses one dry battery as a power source (cassette tape player, etc.), the voltage of the dry battery usually drops from 1.5v to around 1v due to wear and tear. The design of the circuit part is performed as follows.

Among them, LEDs (especially when the forward voltage is 1.4
v) is incorporated in the drive circuit.

As this booster circuit, there is a method utilizing the back electromotive force of the coil shown in FIG. This is because the switch S1 is turned on by the control circuit 1, and the current flowing from the input terminal 2 flows to the ground via the coil L and the switch S1, and the coil L
The energy voltage released from the coil L when the switch S1 is turned off next time is stored in a diode.
The capacitor C1 is charged via D1 and output to the output terminal 3.

There is also a method using a charge pump as shown in FIG. This is because the oscillation circuit 4 first switches
S2 is turned off, switch S3 is turned on, and current flowing from input terminal 5 is supplied to diode D2, capacitor C2 and switch S3.
Through the capacitor to ground and connect the capacitor C3 to the input terminal 5.
, The switch S3 is turned off, the switch S2 is turned on, and the voltage of the input terminal 5 is applied to the capacitor C2 via the diode D3.
And outputs a voltage twice as high as the voltage applied to the input terminal 5.

[Problems to be solved by the invention]

However, in the circuit shown in FIG. 2, the control circuit 1 is required, which complicates the circuit. In the circuit shown in FIG. 3, the diodes D2 and D3 consume the forward voltage (≒ 0.7 V). In the case of low voltage input, there is a problem that this circuit must be configured in a plurality of stages and a plurality of capacitors must be used.

The present invention has been made in view of such a point,
Its purpose is to have a circuit that can boost the voltage simply and efficiently
The purpose is to provide an LED drive circuit.

[Means for solving the problem]

For this purpose, the present invention provides an LED driving circuit for driving an LED by a voltage of a capacitor having one end connected to an output terminal, wherein a first terminal connected to an input terminal and the one end of the capacitor is provided.
, A second transistor connected to the ground side and the other end of the capacitor, a third transistor connected to the input terminal and the other end of the capacitor, and a voltage of the output terminal is lower than a predetermined value. A fourth transistor that turns on the first and second transistors when it is high and turns on the third transistor when it is high.

〔Example〕

Hereinafter, examples of the present invention will be described. FIG. 1 is a diagram showing a circuit of one embodiment. 7 is an input terminal, 8 is an output terminal, and 9 is an LED. Q1 sets the output voltage Vo to R12,
The voltage divided by R13 is applied to the base and its output voltage Vo
Is a transistor (fourth transistor) that is turned on when the voltage becomes equal to or lower than a predetermined voltage. Q2 is a transistor that turns on when the transistor Q1 turns on, and turns on the transistors Q3 and Q4. Q5 is a transistor that turns on when the transistor Q4 turns off and turns on the transistor Q8 (first transistor).
Q6 is also a transistor (second transistor) that turns on when transistor Q4 turns off and drops one end of capacitor C3 to ground potential. Q7 is almost equal to the input voltage at one end of the capacitor C3 which is turned on when the transistor Q3 is
This is a transistor for raising Vi (third transistor). Transistor Q8 serves to charge capacitor C3.

Now, when the output voltage Vo of the output terminal 8 is zero, the base potential of the transistor Q1 has dropped, so that the transistor Q1 turns on and the transistor Q2 turns on. Therefore, transistor Q3 turns off and transistor Q7 turns off, and transistor Q4 turns off and transistors Q5 and Q6
Turns on. Then, when the transistor Q5 is turned on, the transistor Q8 is turned on.

As a result, the current flowing from the input terminal 7 flows through the transistor Q8, the capacitor C3, and the transistor Q6, and the capacitor C3 is charged. That is, _assuming that the voltages between the collector and the emitter of the transistors Q6 and Q8 are V _CE6 and V _CE8 , respectively, the voltage V _{C3 of} this capacitor is charged with the following target: V _C3 = Vi−V _CE6 −V _CE8 (1) . If the battery is fully charged, the output voltage Vo at that time is as follows: Vo = V _C3 + V _CE6 = Vi−V _CE8 (2)

Here, when the output voltage Vo reaches a value that satisfies the above expression (2), the resistors R12 and R12 are turned off so that the transistor Q1 is turned off.
If the division ratio is set by 13, when the equation (2) is satisfied, the transistor Q2 is turned off,
The transistors Q3 and Q4 are turned on. Turning on the transistor Q3 turns on the transistor Q7, and turning on the transistor Q4 turns off the transistors Q5, Q6 and Q8.

As a result, the input terminal 7 is connected to the ground side terminal of the capacitor C3.
From the transistor Va through the transistor Q7 and the output voltage
Will raise Vo. At this time, _assuming that the voltage between the collector and the emitter of the transistor Q7 is _VCE7 , the applied voltage Va is Va = Vi− _VCE7 (3). Accordingly, the output voltage Vo is expressed as Vo = (Vi−V _CE6 −V _CE8 ) + (Vi−V _CE7 ) = 2Vi−3V _CE (4) from the equation (3) and the equation (2). . However, V _CE = V _CE6 = V _CE7 = V _CE8 .

Since this voltage Vo is applied to the LED 9 from the output terminal 8 and consumed, if the voltage drops below the voltage shown in the equation (2), the transistor Q1 is turned on again to charge the capacitor C3, and thereafter (2) When the voltage shown in the formula is reached, the voltage of the capacitor C3 is raised, and this is repeated. Such oscillation continues to drive the LED 9 continuously.

The on-collector-emitter voltage V _CE is, for example, 0.2
It is about V, and the above equation (4) becomes Vo = 2Vi−0.6 (5). Therefore, even if the input voltage Vi is a low voltage of about 1 v, the output voltage Vo can be boosted to about 1.4 v,
A voltage sufficient to drive the LED 9 is obtained.

[Effects of the Invention] As described above, according to the present invention, an input voltage can be boosted with high efficiency, and LE can be increased at an input voltage as low as about 1 v.
There is an advantage that D can be driven reliably.

[Brief description of the drawings]

FIG. 1 is a circuit diagram of an LED drive circuit according to an embodiment of the present invention, and FIGS. 2 and 3 are circuit diagrams of a conventional booster circuit. 1 ... control circuit, 2 ... input terminal, 3 ... output terminal, 4
…… oscillator circuit, 5 …… input terminal, 6 …… output terminal, 7…
... input terminal, 8 ... output terminal, 9 ... LED.

Claims (1)

(57) [Claims] 1. An LED driving circuit for driving an LED by a voltage of a capacitor having one end connected to an output terminal, comprising: an input terminal; a first transistor connected to the one end of the capacitor; A second transistor connected to the other end, a third transistor connected to the input terminal and the other end of the capacitor, and the first and second transistors turned on when a voltage at the output terminal is lower than a predetermined value. And a fourth transistor for turning on the third transistor when high.